Question

In the figure, accelerations of blocks $A$, $B$ and $C$ are indicated by $a$, $b$, and $c$ respectively. Acceleration $b$ and $c$ are w.r.t to the ground. Find the acceleration of the body $A$ in terms of $b$ and $c$ with respect to the ground if the surface is smooth and pulley and string is ideal.

• A. $\sqrt{(b+c{)}^2+a^2}$
• B. $c-(a+b)\cos \theta$
• C. $\sqrt{(b+c{)}^2+c^2-2(b+c).c.\cos \theta }$
• D. $\sqrt{(b+c{)}^2+c^2+2(b+c).c.\cos \theta }$

Answer 1

The correct option is C $\sqrt{(b+c{)}^2+c^2-2(b+c).c.\cos \theta }$

The given system can be reframed as shown

Since the wedge $C$ is moving rightwards, therefore block $A$ will be having two acceleration i.e. one down the inclined plane and other perpendicular to the inclined plane.
From Wedge constraint, the acceleration along the ${\perp }^{ar}$ to the surface of contact should be equal.
$\Rightarrow a_y=c\sin \theta \dots \dots \left(1\right)$

From pulley mounted on wedge + block constraint, we have
${\stackrel{..}{L}}_1+{\stackrel{..}{L}}_2=0\Rightarrow a_x+c\cos \theta -c-b=0$
$\Rightarrow a_x=(b+c)-c\cos \theta \dots \dots \left(2\right)$
And, we know that
$a=\sqrt{a_x^2+a_y^2}=\sqrt{{\left[\right(b+c)-c\cos \theta ]}^2+c^2{\sin }^2\theta }$
$=\sqrt{(b+c{)}^2+c^2-2(b+c)c\cos \theta }$